Work and Energy involving velocity

[AfroQueen]

Homework Statement

A hand exerts a constant horizontal force on a block thatis free to slide on a frictionless surface. The block starts fromrest at point A, and by the time it has traveled a distance d topoint B it is traveling with speed Vb. When the block has traveledanother distance d to point C, will its speed be greater than, lessthan, or equal to 2Vb? Explain your reasoning

Homework Equations

Vi^2=Vo^2+2ad
a= acceleration
d=displacement
Vi= final velocity
Vo= initial velocity

The Attempt at a Solution

i think it would be less but the book says it would be greater than?
So i tried to Vb^2=Vo^2+2ad
Vo=0
Vb= sqrt2ad
then substituted it back into find the velocity at point C which I called Vc but my work keeps giving me a value that is less
Vc^2=Vb^2+2ad
Is there something wrong with my work or strategy

 

[PhanthomJay]

You can also attack this problem using work-energy, as your problem title implies...but by any method, you are correct, its speed at C is less tha twice its speed at B. What did you get for its speed at C compared its speed at B?

 

[AfroQueen]

You can also attack this problem using work-energy, as your problem title implies...but by any method, you are correct, its speed at C is less tha twice its speed at B. What did you get for its speed at C compared its speed at B?

i got Vc= sqrt4ad

 

[PhanthomJay]

Yes, or Vc = 2 sqrt(ad). But you must also determine Vb, to make the comparison. Then Vc/Vb = ??

 

[CxStrike]

Yes, or Vc = 2 sqrt(ad). But you must also determine Vb, to make the comparison. Then Vc/Vb = ??

I have to agree with you. Vc is 1.41 (sqrt2) times greater than Vb.
I did it with both kinematics and work/energy and got the same answer. Unless I'm missing something, the book is wrong.